Question: The equation of hyperbola $H$ is $\dfrac {(y+2)^{2}}{9}-\dfrac{x^2}{36} = 1$. What are the asymptotes?
We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y+2)^{2}}{9} = 1 + \dfrac{x^2}{36}$ Multiply both sides of the equation by $9$ $(y+2)^{2} = { 9 + \dfrac{ x^2 \cdot 9 }{36}}$ Take the square root of both sides. $\sqrt{(y+2)^{2}} = \pm \sqrt { 9 + \dfrac{ x^2 \cdot 9 }{36}}$ $ y + 2 = \pm \sqrt { 9 + \dfrac{ x^2 \cdot 9 }{36}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y + 2 \approx \pm \sqrt {\dfrac{ x^2 \cdot 9 }{36}}$ $y + 2 \approx \pm \left(\dfrac{3 \cdot (x)}{6}\right)$ Subtract $2$ from both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{1}{2}(x) -2$